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Complex valued functions

Let LL be defined by L(y)=y′′+ay′+byL(y)=y\prime\prime+ay\prime+by for a,b∈Ra,b \in \mathbb{R} constants.

Let RR be a complex valued function so that R(x)=P(x)+iQ(x)R(x)=P(x)+iQ(x) for functions P(x)P(x) and Q(x)Q(x). Prove the function f(x)=u(x)+iv(x)f(x)=u(x)+iv(x) satisfies the differential equation L(y)=R(x)x∈IL(y)=R(x) \enspace x \in I if and only if uu and vv satisfy the differential equations L(u)=P(x)L(v)=Q(x)x∈IL(u)=P(x) \enspace L(v)=Q(x) \enspace x \in I.

I literally don't even know where to start, are there some formula or trick that can help?
(edited 7 years ago)
Original post by AishaGirl
Let RR be a complex valued function so that R(x)+iQ(x)R(x)+iQ(x) for functions P(x)P(x) and Q(x)Q(x).


I may be wrong or misreading something here, but should R(x)+iQ(x)R(x)+iQ(x) read R(x)=P(x)+iQ(x)R(x) = P(x)+iQ(x)?
Original post by crashMATHS
I may be wrong or misreading something here, but should R(x)+iQ(x)R(x)+iQ(x) read R(x)=P(x)+iQ(x)R(x) = P(x)+iQ(x)?


Yes you are right, sorry it was a typo I fixed it now.
Original post by AishaGirl
Let LL be defined by L(y)=y′′+ay′+byL(y)=y\prime\prime+ay\prime+by for a,b∈Ra,b \in \mathbb{R} constants.

Let RR be a complex valued function so that R(x)=P(x)+iQ(x)R(x)=P(x)+iQ(x) for functions P(x)P(x) and Q(x)Q(x). Prove the function f(x)=u(x)+iv(x)f(x)=u(x)+iv(x) satisfies the differential equation L(y)=R(x)x∈IL(y)=R(x) \enspace x \in I if and only if uu and vv satisfy the differential equations L(u)=P(x)L(v)=Q(x)x∈IL(u)=P(x) \enspace L(v)=Q(x) \enspace x \in I.

I literally don't even know where to start, are there some formula or trick that can help?


Hint: a+ib=c+ida+ib=c+id if and only if a=c,b=da=c,b=d
ok so I think I have an idea how to start.

If I assume that y=f(x)=u(x)+iv(x)y=f(x)=u(x)+iv(x) satisfies the differential equation then can I say something like

L(y)=R(x)  ⟹  f′′(x)+af′(x)+bf(x)=P(x)+iQ(x)L(y)=R(x) \implies f\prime\prime(x)+af\prime(x)+bf(x)=P(x)+iQ(x)

  ⟹  u′′(x)+iv′′(x)+au′(x)+iav′(x)+bu′(x)+bv′(x)=P(x)+iQ(x)\implies u\prime\prime(x)+iv\prime\prime(x)+au \prime(x)+iav\prime(x)+bu\prime(x)+bv\prime(x)=P(x)+iQ(x)

Is the correct way to go?
Original post by AishaGirl

Is the correct way to go?


Yes. Now collect real and imaginary parts.
Original post by atsruser
Yes. Now collect real and imaginary parts.


Like this?

u′′(x)+au′(x)+bu(x)=P(x)u\prime\prime(x)+au\prime(x)+bu(x)=P(x)

v′′(x)+av′(x)+bv(x)=Q(x)v\prime\prime(x)+av\prime(x)+bv(x)=Q(x)

So L(u)=P(x)L(u)=P(x) and L(v)=Q(x)L(v)=Q(x) ?

Looks good so far?
I think I have proved it now.

Assume L(u)=P(x)L(u)=P(x) and L(v)=Q(x)L(v)=Q(x) for all x∈Ix\in I then

u′′(x)+au′(x)+bu(x)+i(v′′(x)+av′(x)+bv(x))=P(x)+iQ(x)u\prime\prime(x)+au \prime(x)+bu(x)+i(v\prime\prime(x)+av \prime(x)+bv(x))=P(x)+iQ(x)

  ⟹  f′′(x)+af′(x)+bf(x)=R(x)  ⟹  L(y)=R(x) \implies f\prime\prime(x)+af\prime(x)+bf(x)=R(x) \implies L(y)=R(x) for all x∈Ix\in I

Is this correct? It's quite the mouthful :K:

@atsruser @alow @RDKGames @DFranklin can someone tell me if it's correct? :biggrin:
(edited 7 years ago)
Original post by AishaGirl
I think I have proved it now.

Assume L(u)=P(x)L(u)=P(x) and L(v)=Q(x)L(v)=Q(x) for all x∈Ix\in I then

u′′(x)+au′(x)+bu(x)+i(v′′(x)+av′(x)+bv(x))=P(x)+iQ(x)u\prime\prime(x)+au \prime(x)+bu(x)+i(v\prime\prime(x)+av \prime(x)+bv(x))=P(x)+iQ(x)

  ⟹  f′′(x)+af′(x)+bf(x)=R(x)  ⟹  L(y)=R(x) \implies f\prime\prime(x)+af\prime(x)+bf(x)=R(x) \implies L(y)=R(x) for all x∈Ix\in I

Is this correct? It's quite the mouthful :K:

@atsruser @alow @RDKGames @DFranklin can someone tell me if it's correct? :biggrin:


Looks good
Original post by crashMATHS
Looks good


Cheers!
Original post by AishaGirl
I think I have proved it now.

Assume L(u)=P(x)L(u)=P(x) and L(v)=Q(x)L(v)=Q(x) for all x∈Ix\in I then

u′′(x)+au′(x)+bu(x)+i(v′′(x)+av′(x)+bv(x))=P(x)+iQ(x)u\prime\prime(x)+au \prime(x)+bu(x)+i(v\prime\prime(x)+av \prime(x)+bv(x))=P(x)+iQ(x)

  ⟹  f′′(x)+af′(x)+bf(x)=R(x)  ⟹  L(y)=R(x) \implies f\prime\prime(x)+af\prime(x)+bf(x)=R(x) \implies L(y)=R(x) for all x∈Ix\in I


You're halfway there. The question is an "if and only if" proof. Suppose A and B are statements, and A is true if and only if B is true - that means that A implies B and that B implies A i.e.

A⇒BA \Rightarrow B and B⇒AB \Rightarrow A or equivalently A⇔BA \Leftrightarrow B

You have proved the "if" part i.e. A⇒BA \Rightarrow B part. You still need to prove the "only if" part i.e. B⇒AB \Rightarrow A. To do so, you need to show that if f(x)f(x) satisfies the given DE, then L(u)=P(x),L(v)=Q(x)L(u) = P(x), L(v)=Q(x)
Original post by atsruser
You're halfway there. The question is an "if and only if" proof. Suppose A and B are statements, and A is true if and only if B is true - that means that A implies B and that B implies A i.e.

A⇒BA \Rightarrow B and B⇒AB \Rightarrow A or equivalently A⇔BA \Leftrightarrow B

You have proved the "if" part i.e. A⇒BA \Rightarrow B part. You still need to prove the "only if" part i.e. B⇒AB \Rightarrow A. To do so, you need to show that if f(x)f(x) satisfies the given DE, then L(u)=P(x),L(v)=Q(x)L(u) = P(x), L(v)=Q(x)


I believe she's done both directions on two separate posts

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